How do you solve $log3x=log2+log (x+5)$ ?

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Answer 1 · 2016-04-08T10:58:51

$x = 10$

Rearrange slightly to get $x$ on one side and constant on the other

$log3x = log 2 + log(x + 5)$
$log3 + logx = log 2 + log(x+5)$
$logx - log(x+5) = log2 - log3$

Now, using laws of logarithms, make it so you only have a single logarithm on either side

$logx - log(x+5) = log(x/(x+5))$
$log2 - log3 = log(2/3)$

So you have

$log(x/(x+5)) = log(2/3)$

Raise both sides to the power $e$ to cancel out the logarithms (assuming we are dealing with $log_e$ or $ln$ ),

$x/(x+5) = 2/3$

From which we get

$x/(x+5) = 2/3$
$3x = 2(x+5)$
$3x = 2x + 10$
$3x - 2x = 10$
$x = 10$

How do you solve log3x=log2+log (x+5)log3x=log2+log (x+5)?